用C语言程序编写:1+1/3!+1/5!+........+1/n!

Answer 1

#include <iostream>
using namespace std;
int fun(int n)
{
int i,c;
for (c=1,i=2;i<=n;++i)
c*=i;
return c;
}

int main()
{
int i,n;
double sum;
scanf("%d",&n);
for (sum=0,i=1;i<=n;i+=2)
sum+=1.0/fun(i);
printf("%f",sum);
return 0;
}

Answer 2

#include "stdio.h"

double foo(int n)
{
int i;
double rtn = 1.0;
double tmp = 1.0; // 保存上一次的1/n!的值
for (i = 3; i <= n; i += 2)
{
tmp *= (1.0 / i) * (1.0 / (i - 1));
rtn += tmp;
}
return rtn;
}

void main()
{
int n;
scanf("%d", &n);
if (n % 2 == 0)
printf("input error");
else
printf("%.10f\n", foo(n)); // 10表示小数的长度，可修改
}

Answer 3

#include<stdio.h>
void main()
{
int n;
double sum;
printf("请输入 n:");
while(scanf("%d",&n)==1)//可以不断的输入N的值
{
sum=0;
if(n==0)
break;//当n=0.退出
for(int i=1;i<=n;i++)
{
sum=sum+1.0/(i*2-1);//计算和
}
printf("sum=%.6f\n",sum);
}
}

Answer 4

int sum=0,temp=1,i,j;
for(i=1;i<=n;i=i+2)
{
for(j=1;j<=i;j=j+2)
{
temp=temp*j;}
sum=sum+(1/temp);}
大概是这样吧。我没检查